help-33

[trim quest]

You are given expressions for radius and height. All you can really do is plug them in and simplify

to get expressions for SA and V in terms of x

[untold glacier]

Ok I see. and for the SA would I square it since there is a top and bottom?

@trim quest

[trim quest]

Nope, they gave you the whole formula

[untold glacier]

Oh ok

I see

[trim quest]

The top and bottom faces are circles, which have area pi*r^2. The 2 in front of the first term already doubles it for you

[untold glacier]

I apologize but just to confirm i’m doing this correctly, Abd then i would simplify this?

[trim quest]

,rotate

[trim quest]

Looks great 👍

[untold glacier]

Ok Thank you for the help. Im gonna simplify now

[marsh citrus]

@untold glacier Has your question been resolved?

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[vapid nexus]

can i get some pointers here? im not even sure how to interpret this question

[ornate ginkgo]

first off, what is the maximum value that the integrand can be on your interval? that's what your M means

[vapid nexus]

2 / e^(3/2)

[ornate ginkgo]

are you sure?

the maximum value

$2e^{-x/2}$ is a decreasing function as x gets bigger, is it not?

[elfin berry]

blanket

blanket

[vapid nexus]

yes

[ornate ginkgo]

right

so you want to pick the left most values to guarantee the maximum value on your interval right

and the left most value is 0, so what is 2e^{-x/2} when x =0?

[vapid nexus]

which would be 6

oh the left

ahh

i was plugging in 6, made an error

[ornate ginkgo]

yep

[vapid nexus]

would be 2

[trim quest]

Isn't M the maximum of the second derivative of the integrand?

[ornate ginkgo]

oh i couldnt read

[trim quest]

lol

[ornate ginkgo]

nvm, so $f''(x) = \frac18e^{-x/4}$ then

[vapid nexus]

-x/4

not -x/2

[elfin berry]

blanket

[ornate ginkgo]

woops

well whats the maximum value on this between 0 and 6?

[vapid nexus]

1/8

[ornate ginkgo]

right

so lets use that formula now where $M = 1/8, b = 6, a = 0, N = 20$

[elfin berry]

blanket

[ornate ginkgo]

then that should be your answer

[vapid nexus]

oh much simpler than i thought

i appreciate the help! @ornate ginkgo

[ornate ginkgo]

yeah np

[vapid nexus]

.close

[marsh citrus]

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[marsh citrus]

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[stark field]

Hello

[stark field]

integrals have no quotient rule?

its just dx f / dx g ?

[marsh citrus]

@stark field Has your question been resolved?

[main idol]

No

[main idol]

Integrate both sides of the derivative of a quotient could be a rule

Same as integration by parts

[marsh citrus]

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[still temple]

how does the (-1)^m come about? (m is a summation from 0 to infinty) and (z = 1/2+m)

[still temple]

I know it has to do something with the sin but how?

[dark yacht]

For undergrad stuff you'd be better asking in the university section

[still temple]

Alright

.close

[marsh citrus]

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[still temple]

.reopen

[marsh citrus]

✅

[dark yacht]

Just as a fact

Look at the graph of sin(m*pi + pi/2)

If you have an integer m

If you want hard proof use Euler's formula

[still temple]

Which one

[dark yacht]

$e^{(i*\x)} = \cos(x) + i*\sin(x)$

[elfin berry]

JJP
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dark yacht]

Third times the charm

Or not

Anyway, you get it

[still temple]

can I look into the formula?

And get back to you

[dark yacht]

Sure :)

[still temple]

Alright do you want me to dm you or just message here

[dark yacht]

Just message here

[still temple]

Alright

[dark yacht]

Going back to work now but if you get stuck this should help

https://math.stackexchange.com/questions/911716/why-sinn-pi-0-and-cosn-pi-1n

You could also plug sin(m*pi+pi/2) into desmos and convince yourself of it :)

[still temple]

Alright thank you!

[dark yacht]

This proof is for cos(m x pi) but sin(m x pi + pi/2) = cos(m x pi) anyway :)

Feel free to convince yourself of that too with just some basic trig haha

[still temple]

Lol thank you! I really appreciate it!

[dark yacht]

Np!

[still temple]

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[supple oyster]

Can you solve these

[supple oyster]

?

[muted nest]

Translate

[marsh citrus]

@supple oyster Has your question been resolved?

[supple oyster]

No

[supple oyster]

You have to solve this puzzle (kenken)

In the first one you have to write all the number from 1-8 to squares to get the things by doing the actions that are on the corners of black bordered areas but you can only use a number once in every line

The second one is similar to but you can use numbers from 1 to 9

too*

[mystic jetty]

https://tirl.org/software/kenken/

[marsh citrus]

@supple oyster Has your question been resolved?

[marsh citrus]

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[fervent vale]

[Translated task]
In dependence on parameter p/=1, solve the system of equations. If you don't suceed for general p, solve for p=2 (half points)

Can someone help me with Gaussian elimination

The x just not possible it would be like that

Thank you!!

[marsh citrus]

@fervent vale Has your question been resolved?

[marsh citrus]

@fervent vale Has your question been resolved?

[marsh citrus]

@fervent vale Has your question been resolved?

[marsh citrus]

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[timber sky]

Hey i have a question regarding function sequences

[timber sky]

Can the simple convergence —> +infinity on an interval ?

.close

[marsh citrus]

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[still temple]

What do I do next here to simplify?

[torn crest]

rewrite the numerator to be one term

[marsh citrus]

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[rotund walrus]

i dont get what to do here

[proud ice]

it's a problem of conservation of energy

278.32 J is the amount of energy to move the box

Since it's being lifted, I assume it's moving straight up

Thus, you're converting the work into gravitational potential energy

[elfin berry]

~Martin

[rotund walrus]

ohhh

ok

then what

@proud ice

[proud ice]

[rotund walrus]

but

thats not the answer

[proud ice]

No, but it's what you need to use

[rotund walrus]

?

[proud ice]

you know the energy spent

and the mass

[rotund walrus]

yea

[proud ice]

and acceleration due to gravity

[rotund walrus]

yea

[proud ice]

So what's missing?

[rotund walrus]

displacement

so

[proud ice]

This may be a confusing thing to say since the box is not accelerating. I would have been more correct to say "you know mg is the potential force field through which the box is moving"

[rotund walrus]

yea

soooo

whats next

@proud ice

yo

[proud ice]

!15m

[marsh citrus]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[proud ice]

And don't yo me

[proud ice]

You have your answer here

Find your displacement

My only worry is that you never use the constant velocity. But I suppose you don't need it. Perhaps you later find it in the problem?

[rotund walrus]

bruh

[proud ice]

gravitational force field is a conserved vector field, so it doesn't matter what speed you go through it. Just where you start and stop

[proud ice]

use this

[rotund walrus]

so what would the equation look like

i got a very big number

now what

[proud ice]

what did you get

[rotund walrus]

38731

[proud ice]

can you show me your work?

That does seem a bit high

[rotund walrus]

i just multiplied the gravity by the work force and the mass

[proud ice]

That's not using the equation that was shown to you tho

E=mgh

[rotund walrus]

whats h and e

i meant h

[proud ice]

If you don't know, then you need to review gravitational potential

[rotund walrus]

ive never learned that

[proud ice]

Then why are you receiving this problem?

[rotund walrus]

todays out first day learning bout energy

idk

why is it hard?

[proud ice]

crash course:

energy is a quantification of how much work an object with mass can

By "work" I mean "ability to move through a resisting force"

[rotund walrus]

so what did u get for the answer

[proud ice]

I don't give answers

[rotund walrus]

bruh

its a multiple choice question

[proud ice]

bruh yourself I'm not cheet sheet

[rotund walrus]

i dont get the question

i need help

[proud ice]

I'm explaining the question

[rotund walrus]

but your saying i gotta review

[proud ice]

you do

You seem to not know anything about potential energy

[rotund walrus]

whats potential energy

i think we went over it

lemme see

[proud ice]

Si te doy un problema de español y no hablas español, entonces diría que necesitas aprender español si quieres responderla.

[rotund walrus]

wait

bruh im not spanish

rascially motivated

💀

[proud ice]

translation: If I gave you spanish problem, and you do not speak spanish, then I would say that you need to learn to speak spanish if you want to answer it.

[rotund walrus]

i could read that tho

i understood it

[proud ice]

I strongly believe you did if you're getting problems for it

[rotund walrus]

yea

ok so

lets refresh

so we know the mass

gravity

and work force

the equation we got for work force was

wp=(fp)(displacemnt)cos(angle)

[proud ice]

looks good

That's the more general equation, yes

for us, wp is our potential energy (the 232. J whatever)

[rotund walrus]

yea

[proud ice]

fp is the force the object is moving through

mg

displacement is what we are solving for

angle is the angle against the force field through which the object is moving

[rotund walrus]

yea

[proud ice]

I'm assuming the object is being lifted straight up, so the angle should be 0

[rotund walrus]

yea

[proud ice]

Solve for displacement and you're done

[rotund walrus]

soo

its

278.32=(14.2)(9.8)cos(0)?

[proud ice]

Your displacement is missing there

[rotund walrus]

where

[proud ice]

here. Where is it?

[rotund walrus]

14.2 and 9.8

[proud ice]

14.2 and 9.8 together make your fp

[rotund walrus]

multiply

[proud ice]

14.2 is your mass, 9.8 is acceleration, so 14*9.8 is force

[rotund walrus]

139.16

[proud ice]

Has no relation to displacement

[rotund walrus]

oh

so what do u do instead

[proud ice]

start here. wp=278.32, fp=9.8*14.2, angle=0

[rotund walrus]

mhm

got it

then what

[proud ice]

solve for displacement

[rotund walrus]

2

ohhhh

i get it

dang

it was so simple

sorry for wasting your time

thanks tho

[proud ice]

my time is meaningless

yw

[rotund walrus]

bye 🙂

.close

[marsh citrus]

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[marsh citrus]

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[storm canyon]

Hi

[marsh citrus]

Please don't occupy multiple help channels.

[bright jay]

.close

[marsh citrus]

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[blissful mauve]

how do i do this problem

[blissful mauve]

1b.

[elfin berry]

Stephen

[blissful mauve]

root x + 2?

[cobalt sentinel]

no,

so $f(x) = \sqrt{x}$, then $x=$?

[elfin berry]

Stephen

[cobalt sentinel]

how do i get from $\sqrt{x}$ to $x$

[elfin berry]

Stephen

[blissful mauve]

square it

the root of x

[cobalt sentinel]

correct

[cobalt sentinel]

so whats the answer to this

in terms of f(x)

[blissful mauve]

the last of your question is kinda throwing me off

do you just mean in terms of function f

[cobalt sentinel]

yes

[blissful mauve]

just root x?

[cobalt sentinel]

hmm

try this example

[elfin berry]

Stephen

[cobalt sentinel]

essentially, solve for 'x'

[blissful mauve]

just

x

[cobalt sentinel]

x= what?

ok let me make it simpler

if y = 2x, then write 'x' in terms of y

isolate 'x' by itself with no numbers or other variables next to it on one side of the equation

[blissful mauve]

y/2 = x

[cobalt sentinel]

good

[cobalt sentinel]

now do the same for this

[blissful mauve]

is it y^2

[cobalt sentinel]

yes

[cobalt sentinel]

now do it for this

[blissful mauve]

just x?

wait no

i've said it before 💀

[cobalt sentinel]

just the same idea u did for this, dont let the f(x) confuse u, just think of it as a 'y'

[blissful mauve]

is the answer to that question

something related to

squaring

x or y

[cobalt sentinel]

u mean x and f(x)?

[blissful mauve]

yes

[cobalt sentinel]

well yea u have to square both sides, just as u did here

[blissful mauve]

so

x = f(x) squared?

[cobalt sentinel]

correct

[elfin berry]

Stephen

[cobalt sentinel]

do u agree?

[blissful mauve]

yes

[cobalt sentinel]

but we want x+2

so what can we do from here to get to x+2

[blissful mauve]

f(x)^2 = x+1

right?

[cobalt sentinel]

whered u get the +1 from

if u do something to one side, u have to do it to the other as well

[blissful mauve]

oh because my question here is f(x) = root x+1

and im writing x+2 in terms of that function

[cobalt sentinel]

yes thats correct, mb i forgot to look back

so yea

[elfin berry]

Stephen

[cobalt sentinel]

now how we do we get to x+2

[blissful mauve]

i havent been taught the methods for solving this

[cobalt sentinel]

theres nothing to really solve for

we just want to end up with x+2 on the left side, and ultimately we will have to manipulate both sides in order to obtain that

[blissful mauve]

add 1 to each side?

[cobalt sentinel]

correc

tt

nice

so whats our final answer

[blissful mauve]

x + 2 = f(x)^2 +1

?

[cobalt sentinel]

yes but i would write it as $[f(x)]^2$ or $f^2(x)$ because otherwise it looks like ur squaring the $x$ and not the $f(x)$

[elfin berry]

Stephen

[blissful mauve]

hmmm okk

thank you for your patience!

[cobalt sentinel]

no problem

have a good one

[blissful mauve]

.close

[marsh citrus]

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[marsh citrus]

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[tender cave]

For this question

[wary bluff]

oh wait did u want this channel?

[tender cave]

The area enclosed is the triangle between the 4 points

[wary bluff]

4?

[tender cave]

oh nah just saw the question

[glass silo]

It’s already your channel!

[wary bluff]

not 3?

[tender cave]

*3

[wary bluff]

that purple line goin thru is tripping me

How do I deal with that I dont remember it and I cant find anythin in my notes either

do I integrate from 0 to 0.5

then 0.5 to 1

and add them

for 3x and x

[tender cave]

?

[wary bluff]

idk 😭 mb

[tender cave]

ah np

So let the length of the red line be h

And the length of the purple line be b

(The length that is between the points)

[wary bluff]

alr

[wary bluff]

purple or blue?

[tender cave]

purple

[wary bluff]

alright

[tender cave]

The area of a triangle is

1/2(b*h)

[wary bluff]

1/2bh

yeye

[tender cave]

so basically

[wary bluff]

oh wait

[tender cave]

to find the length of the lines

I'll use the distance formula

Which is based upon the Pythagorean theorem

The distance between A(x1, y1) and B(x2,y2) is

[wary bluff]

also we alrdy know the base right

[tender cave]

AB = ((x2-x1)^2+(y2-y1))^(1/2)

Sorry if thats a bit confusing

I havent learnt how to use latex formatting yet

[wary bluff]

all good bro

[fathom ridge]

$\Delta = |\frac{1}{2}[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]|$

[elfin berry]

NEONPerseus

[tender cave]

So now you substitute the point of intersection between the purple line and the blue line

[fathom ridge]

There's a formula

[tender cave]

thanks for that

[fathom ridge]

:/

[tender cave]

oh

Forgot about that

[mortal gazelle]

man i feel like my math is easy compared to yours

but im still stumped

[tender cave]

I forgot about that formula lol

[wary bluff]

I see

[tender cave]

Was about to do it the hard way

[fathom ridge]

There's also a determinant, that I don't remember

[wary bluff]

x1 = 0 x2 = 0.5 and x3 = 1 right?

[fathom ridge]

yeah

[wary bluff]

alright lemem test this one sec

this right

why is it in negative

shit did I mess up

oh

absolute value

wow tysm

so whenever

I have a triangle in the middle of my graph

doin a question like this

I can just use this formula right

[marsh citrus]

@wary bluff Has your question been resolved?

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[fierce belfry]

can i say that the bernouli distribution is basically the binomial distribution

[fierce belfry]

what differs the both of them thou

[fathom vale]

yes

bernoulli is binomial at 1?

[fierce belfry]

wait what does that mean lol

like the number of tries = 1?

[fathom vale]

yes

iirc

[fierce belfry]

wait then theres also something like

in a bern dist, the expected number of tries before succedding is 1/P(succeed)

or something

confused

[fathom vale]

https://math.stackexchange.com/questions/838107/what-is-the-difference-and-relationship-between-the-binomial-and-bernoulli-distr

Note that a binomial random variable with parameter 𝑛=1 is equivalent to a Bernoulli random variable, i.e. there is only one trial.

[fierce belfry]

hmm i see

wait could u help me with part b

like i got the answer but im still confused

like idk why i can use bern

[marsh citrus]

@fierce belfry Has your question been resolved?

[fierce belfry]

hm

.close

[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[echo folio]

I need help converting this geometric series..

[slow zenith]

Crazy hand writing

[echo folio]

Yeah my professor be like that..

[stoic saddle]

was the instruction specifically phrased as "Convert this geometric series"?

or was this you attempting to say it in your own words but failing?

[echo folio]

The whole question is

[stoic saddle]

bc this is not a geometric series.

[echo folio]

This

From "since"

[stoic saddle]

okay

yeah

$\sum_{k=0}^{\infty} \frac{x^k}{k!} = e^x$

[echo folio]

If this is not geometric series what is this?

[elfin berry]

Ann

[stoic saddle]

this is the taylor series for e^x

[echo folio]

Ah right that's the name of this

[stoic saddle]

how much calculus do you know

[echo folio]

I completed calc bc 2-3 years ago so I forgot a lot

I just have to be reminded of the things that existed there

[stoic saddle]

well, brush up on the concept of Taylor series

[echo folio]

And didn't get calc lessons for last 2-3 years

[stoic saddle]

and the forms of said series for specific named functions

[echo folio]

So..

That can be expanded like this

And from here..

[stoic saddle]

yes

$\sum_{k=0}^{\infty} \frac{1}{2^k k!}$ is that same series but with $x=1/2$

[elfin berry]

Ann

[stoic saddle]

do you understand this?

[echo folio]

Yeah

Oh

Ah.........

That's why its e^1/2

[slow zenith]

Is this even math?

[echo folio]

We are supposed to memorize these common forms right?

[echo folio]

Yes, statistics

[fathom ridge]

After a point math isn't about numbers

$\int \ln x \dd{x}$

[elfin berry]

NEONPerseus

[fathom ridge]

Where number

[echo folio]

This is a geometric series right

So we can apply that S= 1/x-1

= (1/x)/(1-(1/x))?

[slow zenith]

[stoic saddle]

common taylor series? yeah you'd do good to know them

@slow zenith do you have anything constructive to say that isn't "wtf is this? i have no idea what y'all are talking about but i will choose to intrude on the convo anyway"

[slow zenith]

yes

$d/dx Si(x)^2

[stoic saddle]

@echo folio yes this is a geometric series but you should be careful

[slow zenith]

How do i make it an image

[echo folio]

What should I be careful about

[stoic saddle]

blindly applying the formula 1/a + 1/a^2 + 1/a^3 + ... = 1/(a-1)

@slow zenith if you want to write something in LaTeX, and specifically a single math formula, it should have dollar signs on both ends of it

[slow zenith]

$d/dx Si(x)^2$

[elfin berry]

CJ_crisjaris

[stoic saddle]

what does this have to do with the problem at hand?

[echo folio]

Hmm..........

[quaint hill]

Please don't interrupt help channels.

Apologies for the disruption

[stoic saddle]

your equation reduces to p_0 / (1 - 1/4) = 1

[echo folio]

how....?

Why does that formula not work here

Probably something to do with a<1 right

Sorry for all of this I think I just need a review of this entire geometric series thing

Do you recommend any resources to look into

[marsh citrus]

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[outer garden]

This is the first time i am seeing this problem. How do i go about summation inside an integral? I could not find any tutorials on youtube or in my textbook

[stark trail]

does that sum converge to a particular value?

[outer garden]

it looks like it

should i simplify the sum first?

[stark trail]

if possible yes

[obsidian shard]

That is the taylor expansion for e^x

[outer garden]

oh i see

will i have to remember this

[stark trail]

I think just probably be able to recognize that it is an expansion of some function

then you can figure out what the function is

[obsidian shard]

This exercise might be helpful to see that the integration of e^x is e^x

[outer garden]

oh

[obsidian shard]

basically, you'll get the same thing you started with

[outer garden]

i see

so it would be e^5-e?

[stark trail]

I think so

[outer garden]

alright thanks

.close

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[stuck marsh]

Hello, we take 4 balls (numbered from 1 to 8) with replacelemnt.
1- How many possibilities do we have?
2- How many possibilities to get 4 different numbers?3 different numbers? 2 different numbers?

For question 1, I said 8^4
Question 2, for the 4 different numbers, I said 8 x 7 x 6 x 5

3 different nubers i said 8 x 7 x 6C2

2 different nulbers i said 8 x 7C3

[stuck marsh]

?

.close

[marsh citrus]

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[chilly quiver]

can someone please help on how to go about solving this

[tight furnace]

you could find the equation of the line

[runic temple]

i think thats the easiest way

[obsidian shard]

Equate the slopes of line AB and BC

[chilly quiver]

would that be y=-5/6x+7/2?

[tight furnace]

Yeah

[chilly quiver]

after that what would i do to find t?

[tight furnace]

The x-value of your point is -3

use that with the line equation

[chilly quiver]

so would y be 6?

[tight furnace]

Yeah

[marsh citrus]

@chilly quiver Has your question been resolved?

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[rose mica]

any hints?

[sleek lake]

uh

like suppose n is 4, so there's 8 ways

and supposedly it involves 3 binary choices

so try to write out all 8

idk

i don't think that helped you just find it impossible to hint

[rose mica]

thanks, but I just found a way

used Power set of {1,2..n}

[sleek lake]

oh i see

[rose mica]

then any subset of it, we place into gaps, to create partitions

[sleek lake]

no i don't

{1,2,3,4}

[rose mica]

oops

[sleek lake]

so {2,3} is a subset

[rose mica]

i meant P{1,2,..n-1}

[sleek lake]

right

so what's {2}, what's {1,2} and what's {1,3}

i'm lost

[rose mica]

if $a\in S \subseteq {1\ldots n-1}$

[sleek lake]

no nvm i see it

[elfin berry]

clemenson

[rose mica]

then place a between a and a+1

[sleek lake]

{2} is 2+2
{1,2} is 1+1+2
{1,3} is 1+2+1

[rose mica]

in the sequence of dots/stars of size n

[rose mica]

yea

[sleek lake]

my way was like, you just count to n and each time you either increase the last term or you add ... + 1, and you don't have this choice at the first step, so n-1

but that's probably cringe

thanks

[rose mica]

didn't understand

but ohwell, I gotta get back to finishing my assignment, cya

.close

[marsh citrus]

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[fervent vale]

How do I do this derivate please

[knotty trellis]

derivative of that?

[proud ice]

Consider chain rule

[knotty trellis]

Firstly, I would start with this

then apply chain rule

[fervent vale]

where did the -1 went please

oh, you put the minus there

[knotty trellis]

$\left(x^{a}\right)^{b}=x^{ab}$

[elfin berry]

MathIsAlwaysRight

[fervent vale]

Thank you!

.close

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[still temple]

how does the (2^m)(m!) become (2m)!!

[still temple]

M is a non negative integer

[marsh peak]

By definition,

(2m)!! = 2m * 2(m - 1) * 2(m - 2) * ... * 2 * 1 = 2^m * m * (m - 1) * (m - 2) * ... * 2 * 1 = 2^m * m!

[marsh citrus]

@still temple Has your question been resolved?

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[still temple]

.reopen

[marsh citrus]

✅

[still temple]

How does the exponent become 2m?

That’s the only question I have

[marsh peak]

There are m amount of 2's

Hence it's 2^m

[still temple]

The series becomes 2m* 2(m-1)…. , I’m just questioning this

[still temple]

Is this because of this?

[marsh peak]

Yes

[still temple]

Alright thank you!

I appreciate your response and have a nice day

[marsh peak]

Likewise

[still temple]

.,¥+|}

.close

[marsh citrus]

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[still temple]

• lol

[marsh citrus]

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[wraith trench]

hello, how can I solve this?

[main idol]

,tex .unit circle

[elfin berry]

riemann

[wraith trench]

but am I supposed to remember all the values?

[still temple]

$\cos(\frac{3\pi}{4}) = \cos(\frac{\pi}{2}+\frac{\pi}{4})$

[elfin berry]

rbit ✨

[still temple]

maybe writing it this way helps

[wraith trench]

how can I solve it?

[main idol]

That's a question for your teacher

[wraith trench]

I don't have a teacher

so I was hoping you guys could help me out

[still temple]

remember some identities with sin and cos?

[main idol]

Then what are you doing the math for

[wraith trench]

I'm just trying to learn math I didn't get good education from school

[wraith trench]

yes

[still temple]

like, cos(x+y)=?

[wraith trench]

well no I haven't learned that just the ratios between sin, cos to unit circle

but should I just memorize all sin and cos values for these special angles or is there a better way I can solve it?

and is there some trick how to memorize them

[trim pulsar]

a really good way i used to memorize these ratios were using triangles

specifically the 30 60 triangle and the 45 45 triangle

[wraith trench]

Yeah but how can I use them for this? they only go up to 90 degrees

[trim pulsar]

ahhh

[trim pulsar]

for that i usually go about using this strategy

find a reference angle and see how many times u add or deduct some value to get to what the question is asking

and the ratio is always the same

and thats because the sinusoidal functions are oscillating

so at some values of pi 0 or some fraction

[wraith trench]

how do I do this convertion?

[trim pulsar]

ok so for cos 3pi/4

lets dissect it

what values of cosine do we know such that pi is the inner function?

we know cos pi = -1

cos 0 = 1

wait

[wraith trench]

yeah

[trim pulsar]

do u know the cast rule>

?

[wraith trench]

never heard of

[trim pulsar]

ok so this is really really useful tool for finding the angles

[wraith trench]

do you mean rotation about the unit circle?

[trim pulsar]

like dead helpufl

yes

[wraith trench]

yeah i think I know that

[trim pulsar]

exactly lets use that

so given this chart we're given a cosine ratio for 3pi/4

[wraith trench]

yeah I am familiar with that ^^

[trim pulsar]

if we visualize 3pi/4

what its asking for is 135

degrees

that puts it in quadrant 2 S

but then its asking for the cosine of it

so that means it will be a negative value

from here lets let visualize a triangle

the angle between the x axis and the hypotenuse of that triangle will be 180/135

thats 45 degrees

now whats cosine 45 degrees?

[wraith trench]

should it no be 180 - 135?

[trim pulsar]

yup

[wraith trench]

1/2

[trim pulsar]

not exactly

its 1/sqrt 2

[wraith trench]

2x^2 = 1

[trim pulsar]

heres a good chart for reference

anyway the reference angle of the 1/sqrt2 from the cosing 45

but the question is asking for 3pi/4

and that puts us in quadrant 2

so that means that 1/sqrt 2

becomes a negative -1/sqrt2

[wraith trench]

i'm not following how can I figure out cos of 135 degrees?

I'll just try and memorize the unit circle, thanks for the help

.close

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[marsh citrus]

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[sly spire]

I was working through the solutions for a practice test and got a bit confused about this term on the RHS here. Could someone try to explain it?

[nocturne fiber]

oh thats just

multiplying by 1

so we can combine the fractions

[sly spire]

ohh they multiplied (k-1)(k) x 1/1?

[marsh citrus]

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[smoky marsh]

isn't this explanation flawed?

f'(x) does exist at x=0 as e^x when x=0 is 1

and that is its only removeable discontinuity

am i misunderstanding something

[marsh citrus]

@smoky marsh Has your question been resolved?

[smoky marsh]

<@&286206848099549185>

<@&286206848099549185>

i just need help understanding the explanation given for the question

[smoky marsh]

<@&286206848099549185>

[marsh citrus]

@smoky marsh Has your question been resolved?

[odd crest]

Could it be

Nvm i was thinking of something else

[smoky marsh]

😭

yea its confusing

[odd crest]

Can you show the information before?

Before the question

[barren carbon]

ok

[smoky marsh]

<@&286206848099549185>

[still temple]

for part (a) you can use a function y = sin(theta). that way there will be no global maxima or minima, since all the extrema would be local (either 1 or -1).
for part (b) you can differentiate y ( i.e. y' = cos(theta)) to show that the critical points are infinitely many.
However, I'm not sure if the proof is supposed to be for the function that you chose in part (a) or any generalized function

[marsh citrus]

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[marsh citrus]

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[quaint arrow]

I am confused on the first part

[quaint arrow]

I was thinking if x1 = 0, it's obvious that it's a fixed point and thus not really biological relevant

so then I was thinking that the only fixed points that are biologically relevant is x = 1-1/a such that a is on the range (1,3)

[marsh citrus]

@quaint arrow Has your question been resolved?

[quaint arrow]

<@&286206848099549185>

[marsh citrus]

@quaint arrow Has your question been resolved?

[marsh citrus]

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[lean flame]

I am deriving arcsin ( x / sqrt(1+x^2) )

I did d/dx sin y = d/dx x / sqrt(1+x^2) and got to:

cos y dy/dx = 1 / ( (1+x^2) sqrt(1+x^2) )

I do not know where to go after this

[sage pollen]

Derivating* btw.

Anyway, your d/dx of x/sqrt(1+x^2) seems to be incorrect.
The derivative of u/v where u and v are functions of x is (vdu - udv)/v^2

lmk when you fix that @lean flame

[still temple]

Derivating isnt a word either

its differentiating

[sage pollen]

It isn't???

Oh right

Yeah I'm just asleep

[lean flame]

,w derive x / sqrt (1+x^2)

[elfin berry]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Diesel#5243

[sage pollen]

Ok clearly I need to go to bed

[lean flame]

how is it wrong?

[sage pollen]

Yeah I'm sorry about that.

Anyway, to continue with your method, you just divide both sides by cos(y) and then substitute the value of cos(y) from your first equation of sin(y) = x/(1+x^2)^0.5

[sage pollen]

It's not, I miscalculated

My bad

[lean flame]

cos y = sqrt (1 - sin y)?

[sage pollen]

sqrt(1-sin^2(y)) but yes

[lean flame]

oh

[sage pollen]

As you can probably guess, that's a bit complicated, although it will work. An alternative for future instances of similar problems, we have the chain rule:
d/dx f(g(x)) = f'(g(x)) * g'(x)

[lean flame]

why is this shit so complicated 😢

always hated trig

[sage pollen]

I think you'll find that the expression simplifies a bit

but yeah, it's not the best

[sage pollen]

Anyway, chain rule is more useful in the future

[lean flame]

ok I got the correct answer. this was tough

thanks

[sage pollen]

np, sorry for the initial confusion

[lean flame]

.close

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[undone epoch]

Hi

[rigid narwhal]

Hi

[undone epoch]

In 2019 and 2020, we invested 400 euros in the bank at the beginning of each month. We will withdraw the saved funds in the following way: with equal withdrawals of b euros at the beginning of each month in 2022, 2023 and 2024 and with another withdrawal of c = 3,000 euros at the end of 2024. Calculate b if interest is paid in a conformal manner with monthly capitalization and if the annual depreciating interest rate is 3% all the time.

Can you please help me solve it

Nobody solved yet

[proper zodiac]

<@&268886789983436800>

[marsh citrus]

@undone epoch Has your question been resolved?

[undone epoch]

Can you repeat

Please

It somehow got deleted

@proper zodiac

[marsh citrus]

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[still temple]

.close

[marsh citrus]

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[still temple]

Help

[still temple]

Wait nvm I figured it out

.close

[marsh citrus]

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[clear chasm]

how would you go about this question???

[feral rock]

alr

tell me the original volume of the box

[clear chasm]

165

[feral rock]

alr

then you would let say prime factorize 400

because it could be anything from 300-900 cm cubed

so we take 400 as an example and prime factorize it

or you could do 600 which is better

or

[clear chasm]

so with 400 if i prime factorized it 2^4 x 5^2?

[feral rock]

ok

actually the easier way is just multiply each number by the x and see if the result is between 300-900

for example

if x=2

multiply 3 ，5，11 by 2 so it would be 6x10x22
which is beyond 900

you can multiply each number by let say 0.5 or 0.75 or 1.5 or 1.75

[clear chasm]

if i used 1.5 its within 300-900

[feral rock]

i mean is it

[clear chasm]

yeah its 556.875

[feral rock]

then it works

[clear chasm]

so then 1.5 would be the number i sub into the x values to get give me my answer?

[feral rock]

yes

[clear chasm]

okay thanks:)

[feral rock]

1.75 works between ranges 1.3 to 1.75 for x works

[clear chasm]

i wonder if the teacher wants multiple x values tho

oohhhh okay so theres a range or {1.3 ≤ x ≤ 1.75}

[feral rock]

i think so

actually to be exact

[clear chasm]

yeah because 1.8 is over 900

[feral rock]

yeah

[feral rock]

this is the general range

you can be specific or not it really depends on your teacher

[clear chasm]

yeah it says what values of x so im assuming it wants the broad x values that work within 300-900

[feral rock]

alr anything else u need help in？

[clear chasm]

no thats it thanks 🙂

[feral rock]

alr

u wanna close it？

.close

[marsh citrus]

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[eternal tundra]

anyone got some last minute tips for analysis 1?
I believe it is calculus 1 in english

[eternal tundra]

last minute because i had like 0 time to prepare because of other exams 😦

[glass perch]

I just wrote my exam 😂

[eternal tundra]

und, wie wars denn so?

[glass perch]

Good luck

[eternal tundra]

thank you^^

[glass perch]

Na ja, nicht so berauschend. Mal sehen, ob bestanden oder nicht

[eternal tundra]

ich hatte gestern meine experimentalphysik 3 prüfung und morgen analysis 1

[glass perch]

Was ich dir raten kann ist die Voraussetzung zu lernen z.B das du L'hôspital anwenden kannst usw.

[eternal tundra]

die kriegen es echt immer hin die klausuren möglichst dicht zu packen

ableitungen kommen bei uns erst in ana 2 🙂

[glass perch]

Wuuuuuus

[eternal tundra]

wir haben folgen, reihen und son kram bisher nur
aber dafür halt sehr weit ausgeführt schätze ich

[glass perch]

Was ich etwas bereut habe ist die Trig Identitäten nicht aufzuschreiben. Hat viel Zeit gekostet

[eternal tundra]

ich hoffe einfach auf sachen wie:
finde grenzwert von a_n wobei a_n ein bruch von zwei polynomen ist

sowas ist nett 🙂

[glass perch]

Ja

Oder Induktion 😄

[eternal tundra]

ja das ist super

aber ich sehe schon wie der ankommt: sie kennen aus der vorlesung den satz xyz, beweisen sie diesen

[glass perch]

Das ist in der Regel nicht so. Du solltest aber schon den Inhalt der wichtigen Sätze kennen

[eternal tundra]

joa das stimmt wohl

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[glass perch]

Nochmal viel Erfolg morgen. Und vergiss nicht zu schlafen! Ist wichtig, dass du dich morgen konzentrieren kannst

[marsh citrus]

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[upper gust]

How does it make intuitive sense, that

(1+1/x)^x = e when x-> infinity and x-> negative infinity

[still temple]

Without some previous definition of what 'e' is, I don't think you can do much to get it intuitively other than like plugging in numbers and noticing their converging behaviour

It's a very special limit for a reason after all

Like if you have 'e' previously defined, then you can do stuff like L'hopital and other methods to "prove" It, but you really can't figure out that it is e without knowing what e is to begin with

[upper gust]

e for me is defined with the taylor series

[still temple]

There is the compound interest explanation which you can look up though

[upper gust]

and i unserstand the limit for positive infinity

but not for negative inf

[still temple]

Negative infinity it is the same story but rephrased

Like

Man I really don't want to latex on phone

But anyways

[upper gust]

ik its pain D:

[still temple]

[
\lim_{x \to -\infty} \parens{1+\f 1x}^x = \lim_{x \to \infty} \parens{ 1 - \f 1x}^{-x} \
\lim_{x \to \infty} \parens{\f{1}{1-\f 1x}}^x = \lim_{x \to \infty} \parens{\f{x}{x-1}}^x \
= \lim_{x \to \infty} \parens{\f{x+1}{x}}^{x+1} = \lim_{x \to \infty} \parens{1+ \f 1x} \parens{1 + \f 1x}^x = e ]

Dammit okay let's see

[still temple]

@upper gust

Kinda scuffed latex but it works

If you are confused on what I did in the second last step, I basically re-indexed the limit, you would need to have subtract 1 from the thing you are approaching, but infinity -1 is just infinity after all

I believe this is the common explanation?

[upper gust]

thanks!

This helped

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[vestal mist]

i got to this

L = 3/5(K), so KL=K*K*(3/5)=3/5(K^2).

not sure what to do next\

[split spoke]

what have you tried?

[vestal mist]

i tried to find the divisors

but it didnt go so well

[split spoke]

could you write KL in terms of L

[marsh citrus]

@vestal mist Has your question been resolved?

[vestal mist]

sure

KL=5/3*(L^2)

[split spoke]

can 5/3 times a square be a square?

[marsh citrus]

@vestal mist Has your question been resolved?

[vestal mist]

well no

but maybe 5/3*L^2 could

[split spoke]

L is a positive integer

[vestal mist]

yes

and if L has a factor of 3

[split spoke]

then L^2 has a factor of nine

take L = 3

[vestal mist]

yes

[split spoke]

5/3 L^2 = 15

[vestal mist]

mhm

[split spoke]

thats not a square

[vestal mist]

how do we know it cant result in a perfect square

ending in 0

thats only one instance of L

[split spoke]

cause three times a square

is not gonna be a square

same with 5

[vestal mist]

because it will have an odd number of 5s?

[split spoke]

yes

[vestal mist]

i see

thanks

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[still temple]

Hello

Is this solution correct?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[still temple]

Hi

[still temple]

For Stem and Leaf Diagram, Odd Or Even, finding the mean should be the same right? I mean n+1/2

Coz my friend said there are different ways

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[split spoke]

My first idea was to do an oscillating sequence but thats not gonna work

Maybe if a_n is the sum of the first n terms of a cauchy sequence?

hmmm

[main idol]

Try functions that grow slow

But still diverge to infinity

[split spoke]

a_n is the harmonic series?

[nocturne fiber]

sure you can pick that

see if it works

[split spoke]

|a_n+1 - a_n| = |b_n+1| = |1/n|

seems to work

archimedian principle

[smoky siren]

Osculating has a mathematical meaning that is different from what you are trying to write here.

[split spoke]

its with an i

mb

[smoky siren]

It's mostly an aside, but words matter, especially when you replace one with another that is also mathematical terminology

[nocturne fiber]

oscillating means

(-1)^n a_n

[marsh citrus]

@split spoke Has your question been resolved?

[marsh citrus]

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[ripe glen]

I'm using the tabular method to solve this problem but I don't know how approaching the right can work?

[unreal oxide]

0.71, 0.701, 0.7001, 0.70001, 0.700001 or something

u can really take any sequence of numbers

that is close to 0.7